IV. Acids and Bases. react with metals to produce hydrogen gas

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1 IV. Acids and Bases Properties of Acids and Bases Acids ph<7.0 taste sour react with metals to produce hydrogen gas ph paper turns red/orange phenolphthalein colourless bromothymol blue yellow cabbage juice pink Bases ph>7.0 taste bitter feel slippery ph paper turns blue/green phenolphthalein pink bromothymol blue blue cabbage juice blue ph and poh ph ( power of hydrogen ) is a measure of the acidity of a solution: the lower the ph, the more acidic the solution is and the higher the ph, the more basic the solution is poh is a measure of the basicity (ar alkalinity) of a solution: the lower the poh, the more basic the solution is and the higher the poh, the more acidic the solution is ph or poh = 7.0 indicates a neutral solution The ph Scale Arrhenius Concept Acid: substance that will dissociate to give H ions Base: substance that will dissociate to give OH - ions The Ahhrenius concept provides a good description of strong acids and bases Strength of Acids and Bases The strength of acids and bases is a measure of the degree of ionization; a strong acid or base will completely dissociate in water. Strong Acids: HCl H Cl - hydrochloric acid HNO 3 H - NO 3 HBr H Br - hydrobromic acid HIO 4 H - IO 4 HI H I - hydroiodic acid H 2 SO 4 H - HSO 4 HClO 4 H - ClO 4 perchloric acid HClO 3 H - ClO 3 nitric acid periodic acid sulphuric acid chloric acid Strong Bases: LiOH Li OH - lithium hydroxide Ba(OH) 2 Ba 2 2OH - barium hydroxide NaOH Na OH - sodium hydroxide Sr(OH) 2 Sr 2 2OH - strontium hydroxide KOH K OH - potassium hydroxide Ca(OH) 2 Ca 2 2OH - calcium hydroxide RbOH Rb OH - rubidium hydroxide (note: other group II elements are insoluble in water) CsOH Cs OH - cesium hydroxide Calculations for Strong Acids and Bases ph = -log[h ] poh = -log[oh - ] [H ] = 10 -ph [OH - ] = 10 -poh ph poh = x10-14 = [H ] [OH - ]

2 ph, poh, and Significant Figures For ph and poh, only the digits after the decimal place are significant. (ie, ph= has two significant digits since there are two numbers after the decimal place; poh = has three significant digits since there are three numbers after the decimal place) ex. A solution has an [H ] of M, calculate the ph. ex. A solution has an [OH - ] of 6.5x10-5 M, calculate the poh ex. acid rain has a ph of 4.80, calculate the [H ]. ex. ammonia has a poh of 3.78, calculate the [OH - ] ex. calculate the [H ], ph, poh, and [OH - ] of M HCl ex. calculate the [OH - ], poh, ph, and [H ] of M Sr(OH) 2 The Ahhrenius Concept does not adequately describe some compounds that exhibit acidic or basic properties, so a more general definition is required. Bronsted-Lowry Model Acid: a proton (H ) donor Base: a proton (H ) acceptor Acid/Base Equilibria A weak acid or base is one that does not completely dissociate in water; instead it forms an equilibrium with water. Acids ex. HA (aq) H 2 O (l) H 3 O (aq) A - (aq) acid base conjugate conjugate acid base note: the proton is donated by the acid (HA) and accepted by the base (H 2 O) The conjugate acid of water is the hydronium ion (H 3 O ) which has one additional proton and the conjugate base of HA is A - which has one less proton. The equilibrium expression for this reaction can be written as follows: K a [ HO 3 ][ A] = [ HA] Bases ex. B (aq) H 2 O (l) OH (aq) BH (aq) base acid conjugate conjugate base acid Where K a is the acid dissociation constant. note: the proton is accepted by the base (B) and donated by the acid (H 2 O) The conjugate base of water is the hydroxide ion (OH - ) which has one less proton and the conjugate acid of B is BH which has one additional proton. The equilibrium expression for this reaction can be written as follows: K b [ OH ][ BH = [ B] ] Where K b is the base dissociation constant.

3 K a values for Common Acids. The higher the K a value, the stronger the acid. Acid Formula K a iodic acid HIO oxalic acid H 2 C 2 O 4 5.9x10-2 sulphurous acid H 2 SO 3 1.5x10-2 hydrogen sulphate ion - HSO 4 1.2x10-2 chlorous acid HClO 2 1.2x10-2 phosphoric acid H 3 PO 4 7.5x10-3 hydrofluoric acid HF 7.2x10-4 nitrous acid HNO 2 4.0x10-4 hydrogen oxalate ion - HC 2 O 4 6.1x10-5 acetic acid HC 2 H 3 O 2 (or CH 3 COOH) 1.8x10-5 carbonic acid H 2 CO 3 4.3x10-7 hydrogen sulphite ion - HSO 3 1.0x10-7 hydrosulphuric acid H 2 S 1.0x10-7 dihydrogen phosphate ion - H 2 PO 4 6.2x10-8 hypochlorous acid HOCl 3.5x10-8 hypobromous acid HOBr 2.0x10-9 hydrocyanic acid HCN 6.2x10-10 boric acid H 3 BO 3 5.8x10-10 ammonium ion NH 4 5.6x10-10 hydrogen carbonate ion - HCO 3 5.6x10-11 hydrogen phosphate ion 2- HPO 4 4.8x10-13 hydrogen sulphide ion HS - 1.3x10-13 K b values for Common Bases. The higher the K b value, the stronger the base. Base Formula/ Conjugate Acid K b diethylamine (C 2 H 5 ) 2 NH/ (C 2 H 5 ) 2 NH 2 1.3x10-3 ethylamine C 2 H 5 NH 2 / C 2 H 5 NH 3 5.6x10-4 methylamine CH 3 NH 2 / CH 3 NH 3 4.4x10-4 triethylamine (C 2 H 5 ) 3 N/ (C 2 H 5 ) 3 NH 4.0x10-4 ammonia NH 3 / NH 4 1.8x10-5 hydrazine H 2 NNH 2 (or N 2 H 4 )/ H 2 NNH 3 3.0x10-6 hydroxylamine HONH 2 / HONH 3 1.1x10-8 pyridine C 5 H 5 N/ C 5 H 5 NH 1.7x10-9 aniline C 6 H 5 NH 2 / C 6 H 5 NH 3 3.8x10-10 ex. HF (aq) H 2 O (l) H 3 O (aq) F - (aq) acid base conjugate conjugate acid base K a [ HO 3 ][ F = [ HF] ] ex. NH 3 (aq) H 2 O (l) OH (aq) NH 4 (aq) base acid conjugate conjugate base acid K b [ OH ][ NH = [ NH ] 4 3 ] ex. Write a K a /K b expression for the following acid/base equilibria. Identify the conjugate acid-base pairs. (1) HC 2 H 3 O 2 (aq) H 2 O (l) H 3 O (aq) C 2 H 3 O - 2 (aq) (2) CH 3 NH 2 (aq) H 2 O (l) OH - (aq) CH 3 NH 3 (aq) ex. Complete the following acid-base equilibria. Identify the conjugate acid-base pairs and write a K a /K b expression. (1) HOBr (aq) H 2 O (l) (2) H 2 NNH 2 (aq) H 2 O (l)

4 Hydrated Metal Ions Certain metal ions (particularly transition metal ions and aluminum) when in solution will form hydrated complexes that may act as acid. ex. When Al 3 is in solution, it forms [Al(H 2 O) 6 ] 3, the acid equilibrium for this complex is as follows: (K a = 1.4x10-5 ) [Al(H 2 O) 6 ] 3 (aq) H 2 O (l) H 3 O (aq) [Al(H 2 O) 5 OH] 2 (aq) ex. When Fe 3 is in solution, it forms [Fe(H 2 O) 6 ] 3. What is the acid equilibrium for this complex? (K a = 6.0x10-3 ) [Fe(H 2 O) 6 ] 3 (aq) H 2 O (l) Calculations for Weak Acids and Bases The equilibrium concentration of the hydronium ion or hydroxide ion must be determined using an ICE box. ex. Determine the ph and poh of 1.0 M solution of HF (K a = 7.2x10-4 ). ex. Determine the poh and ph of a 1.0 M solution of NH 3 (K b = 1.8x10-5 ). ex. The ph of a M solution of citric acid (H 3 C 6 H 5 O 7 ) is Determine K a. ex. The poh of a M solution of toluidine (CH 3 C 6 H 4 NH 2 ) is Determine K b.

5 Percent Dissociation (also called percent ionization) The percent dissociation of a compound is a measure of the degree of dissociation. The percent dissociation can be calculated according to the following equation: amount dissociated Percent dissociation = x100% amount dissociated = x initial concentration ex. Calculate the percent dissociation for a 0.22 M solution of C 2 H 5 NH 2 (K b = 5.6x10-4 ) ex. A M solution of lactic acid (HC 3 H 5 O 3 ) is 3.7% dissociated at equilibrium. Calculate the K a value for lactic acid. Amphoteric Compounds amphoteric: a compound that can act as either an acid or a base. ex. Water is amphoteric: H 2 O (l) H 2 O (l) H 3 O (aq) OH - (aq) note: the autoionization of water produces both a hydronium ion and a hydroxide ion so the resulting solution is neutral. The equilibrium expression for this reaction is as follows: K = [ HO ][ OH, where K w = 1.0x10-14 (K w is called the dissociation constant of water) w 3 ] other amphoteric substances include: the hydrogen sulphate ion (HSO 4 - ), the hydrogen sulphite ion (HSO 3 - ), the dihydrogen phosphate ion (H 2 PO 4 - ), the hydrogen phosphate ion (HPO 4 2- ), the hydrogen sulphide ion (HS - ) etc. Relationship between K a and K b For a weak acid, if the K a value is known, then the K b value of its conjugate base can be calculated. For a weak base, if the K b value is known, then the K a value for its conjugate acid can be calculated. This relationship is described according to the following equation: K = K K K w = 1.0x10-14, K a is the acid dissociation constant, and K b is the base dissociation constant w a b Strength of a conjugate acid or base The stronger the acid, the weaker its conjugate base and the stronger the base, the weaker its conjugate acid. ex. Which of the following substances is a stronger acid, HF or HCN? Which of the following substances is a stronger base, F - or CN -? HF has K a = 7.2x10-4 and HCN has K a = 6.2x Since K a for HF is greater than K a for HCN, HF is a stronger acid than HCN. F - has K b = 1.0x10-14 /7.2x10-4 = 1.4x10-11 and CN - has K b = 1.0x10-14 /6.2x10-10 = 1.6x10-5. Since K b for CN - is greater than K b for F -, CN - is a stronger base than F -. ex. Which of the following substances is a stronger base, NH 3 or C 5 H 5 N? Which of the following substances is a stronger acid, NH 4 or C 5 H 5 NH? NH 3 has K b = 1.8x10-5 and C 5 H 5 N has Kb = 1.7x10-9. Since K b for NH 3 is greater than K b for C 5 H 5 N, NH 3 is a stronger base than C 5 H 5 N. NH 4 has K a = 5.6x10-10 (given on K a table) and C 5 H 5 NH has K a = 1.0x10-14 /1.7x10-9 = 5.9x10-9. Since K a for C 5 H 5 NH is greater than K a for NH 4, C 5 H 5 NH is a stronger acid than NH 4.

6 Comparing Equilibria for Amphoteric compounds ex. the hydrogen carbonate ion, HCO 3 - can act as an acid (donating a proton) or a base (accepting a proton) as an acid: HCO 3 - (aq) H 2 O (l) H 3 O (aq) CO 2-3 (aq) K a = 5.6x K as a base: HCO - 3 (aq) H 2 O (l) OH - w 1.0x10 (aq) H 2 CO 3 (aq) Kb = = = 2.3x10 7 Ka 4.3x10 Since K b >K a, a solution containing hydrogen carbonate would have a basic ph. 8 Acid Base Reactions Weak acids can react with weak bases and result in an acid/base equilibrium. Consider the following reactions and identify the conjugate acid-base pairs. (1) NH 3 (aq) H 2 S (aq) HS - (aq) NH 4 (aq) Calculating K for an acid-base reaction: Consider the compounds acting as acids for the reaction: H 2 S (aq) H 2 O (l) H 3 O (aq) HS - (aq) K a = 1.0x10-7 NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3 (aq) K a = 5.6x10-10 Note, if the second equilibrium is reversed (and the reciprocal of the equilibrium constant is taken), the two equilibria can add up to give the overall equilibrium (and the K for the equilibrium can be obtained by multiplying K for each of the reactions) H 2 S (aq) H 2 O (l) H 3 O (aq) HS - (aq) K a = 1.0x10-7 H 3 O (aq) NH 3 (aq) NH 4 (aq) H 2 O (l) K = 1/5.6x10-10 = 1.8x10 9 NH 3 (aq) H 2 S (aq) HS - (aq) NH 4 (aq) K =1.0x10-7 x 1.8x10 9 = 1.8x10 2 (note: H 3 O (aq) and H 2 O (l) are cancelled when the equilibria are added) An important result is that K>1. This is significant since H 2 S is a stronger acid than NH 4, therefore the equilibrium position is driven to the right, favouring the products. This same result could be derived by considering the compounds that act as bases in the reaction: NH 3 (aq) H 2 O (l) OH - (aq) NH 4 (aq) K b = 1.8x10-5 HS - (aq) H 2 O (l) OH - (aq) H 2 S (aq) K b = 1.0x10-14 /1.0x10-7 = 1.0x10-7 Note, if the second equilibrium is reversed (and the reciprocal of the equilibrium constant is taken), the two equilibria can add up to give the overall equilibrium (and the K for the equilibrium can be obtained by multiplying K for each of the reactions) NH 3 (aq) H 2 O (l) OH (aq) NH 4 (aq) K b = 1.8x10-5 OH - (aq) H 2 S (aq) HS (aq) H 2 O (l) K = 1/1.0x10-7 = 1.0x10 7 NH 3 (aq) H 2 S (aq) HS - (aq) NH 4 (aq) K =1.8x10-5 x 1.0x10 7 = 1.8x10 2 (as determined above) (note: OH - (aq) and H 2 O (l) are cancelled when the equilibria are added)

7 (2) HOCl (aq) C 6 H 5 NH 2 (aq) OCl - (aq) C 6 H 5 NH 3 (aq) HOCl (aq) H 2 O (l) H 3 O (aq) OCl - (aq) K a =3.5x10-8 C 6 H 5 NH 3 (aq) H 2 O (l) H 3 O (aq) C 6 H 5 NH 2 (aq) K a = 1.0x10-14 /3.8x10-10 = 2.6x10-5 HOCl (aq) H 2 O (l) H 3 O (aq) OCl - (aq) K a =3.5x10-8 C 6 H 5 NH 2 (aq) H 3 O (aq) C 6 H 5 NH 3 (aq) H 2 O (l) K = 1/2.6x10-5 = 3.8x10 4 HOCl (aq) C 6 H 5 NH 2 (aq) OCl - (aq) C 6 H 5 NH 3 (aq) K = 3.5x10-8 x 3.8x10 4 = 1.3x10-3 An important result is that K<1. This is significant since C 6 H 5 NH 3 is a stronger acid than HOCl, therefore the equilibrium position is driven to the left, favouring the reactants. (again, the K value could also be calculated by considering the compounds acting as bases) Mixtures of Acids In general, only the strongest acid will significantly contribute to the concentration of hydronium ions. ex. Calculate the ph of a mixture containing 0.50 M HCl and 0.50 M HF. Since HCl is a strong acid and HF is a weak acid, the HF will not significantly contribute to the H concentration, so the ph can be calculated from the HCl concentration. ex. Calculate the ph of a mixture containing 0.10 M HNO 2 (K a = 4.0x10-4 ) and 0.10 M HCN (K a = 6.2x10-10 ) Since Ka for HNO 2 >> than Ka for HCN, only the HNO 2 will significantly contribute to [H 3 O ]. An ICE table can then be used to determine ph. Mixtures of Bases In general, only the strongest base will significantly contribute to the concentration of hydroxide ions. ex. Calculate the poh of a mixture containing 0.20 M Sr(OH) 2 and 0.20 M NH 3. Since Sr(OH) 2 is a strong base and NH 3 is a weak base, the NH 3 will not significantly contribute to the OH - concentration, so the poh can be calculated from the Sr(OH) 2 concentration. ex. Calculate the poh of a mixture containing 0.50 M C 6 H 5 NH 2 (K b = 3.8x10-10 ) and 0.50 M NH 3 (K b = 1.8x10-5 ) Since Kb for NH 3 >> than Kb for C 6 H 5 NH 2, only the NH 3 will significantly contribute to [OH - ]. An ICE table can then be used to determine poh. Solutions of Acids or Bases Containing a Common Ion of the Conjugate base/acid An ICE table can be used to determine ph/poh, where the initial concentration of the conjugate ion will not be zero. ex. A solution contains 1.0 M HF and 1.0 M NaF, Calculate the ph of the resulting solution. (How does this ph compare to a 1.0 M solution of HF?) ex. A solution contains 1.0 M NH 3 and 1.0 M NH 4 Cl, Calculate the ph of the resulting solution. (How does this ph compare to a 1.0 M solution of NH 3?)

8 Number of Acidic Protons monoprotic acids: have only one acidic hydrogen atoms (ex. HCl and HC 2 H 3 O 2 ) diprotic acids: have two acidic hydrogen atoms (ex. H 2 SO 4 and H 2 C 2 O 4 ) polyprotic acids: have three or more acidic hydrogen atoms (ex. H 3 PO 4 ) ex. Successive ionization of carbonic acid. H 2 CO 3 H 2 O (l) H 3 O (aq) HCO 3 - (aq) K a1 = 4.3x10-7 HCO 3 - H 2 O (l) H 3 O (aq) CO 3 2- (aq) K a2 = 5.6x10-11 H 2 CO 3 2H 2 O (l) 2H 3 O (aq) CO 2-3 (aq) K = K a1 x K a2 = 4.3x10-7 x 5.6x10-11 = 2.4x10-17 For a diprotic acid, K a1 >K a2. (The first proton is more easily lost than the second proton.) Acid-Base Properties of Salts The cation or anion of a salt may hydrolyze (react with water) and produce an acidic or basic solution. Cations Cations found in strong bases will not hydrolyze. Cations that are the conjugate acids of weak bases will hydrolyze to produce acidic solutions. Anions Anions found in strong acids will not hydrolyze. Anions that are the conjugate bases of weak acids will hydrolyze to produce basic solutions. ex. Determine if the ph of the following solutions will be acidic, basic, or neutral. (1) NaCl NaOH is a strong base so Na will not hydrolyze HCl is a strong acid so Cl - will not hydrolyze Since neither the cation or anion hydrolyze, the solution will be neutral (2) NaNO 2 NaOH is a strong base so Na will not hydrolyze NO 2 - is the conjugate base of the weak acid HNO 2, so NO 2 - will hydrolyze: NO 2 - (aq) H 2 O (l) OH - (aq) HNO 2 (aq) The resulting hydrolysis reaction produces hydroxide ions so the solution is basic. (3) NH 4 Cl HCl is a strong acid so Cl - will not hydrolyze NH 4 is the conjugate acid of the weak base NH 3, so NH 4 will hydrolyze: NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3 (aq) The resulting hydrolysis reaction produces hydronium ions so the solution is acidic. (4) NH 4 NO 2 NO 2 - is the conjugate base of the weak acid HNO 2, so NO 2 - will hydrolyze: NO 2 - (aq) H 2 O (l) OH - (aq) HNO 2 (aq) NH 4 is the conjugate acid of the weak base NH 3, so NH 4 will hydrolyze: NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3 (aq) Since both ions hydrolyze, the K a and K b values for the resulting equilibria must be compared. HNO 2 has K a = 4.0x10-4, therefore, NO 2 - has K b = 2.5x10-11 NH 4 has a K a = 5.6x10-10 Since K a > K b, the hydrolysis of NH 4 will favour the products more than the hydrolysis of NO 2 -, as a result there will be more hydronium ions than hydroxide ions and the solution will be acidic.

9 Calculations for Salts ex. Calculate the ph and poh of a 0.20 M solution of NaF. NaOH is a strong base so Na will not hydrolyze F - is the conjugate base of the weak acid HF, so F - will hydrolyze: F - (aq) H 2 O (l) OH - (aq) HF (aq) K b = K w /K a = 1.0x10-14 /7.2x10-4 = 1.4x10-11 ex. Calculate the ph and poh of a M solution of NH 4 Br. HBr is a strong acid so Br - will not hydrolyze NH 4 is the conjugate acid of the weak base NH 3, so NH 4 will hydrolyze: NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3 (aq) K a = 5.6x10-10 ex. Calculate the ph and poh of a M solution of NH 4 NO 2. NO 2 - is the conjugate base of the weak acid HNO 2, so NO 2 - will hydrolyze: NO 2 - (aq) H 2 O (l) OH - (aq) HNO 2 (aq) K b = 2.5x10-11 NH 4 is the conjugate acid of the weak base NH 3, so NH 4 will hydrolyze: NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3 (aq) K a = 5.6x10-10

10 Buffers buffer: is a solution that resists changes in ph. A buffer contains a weak acid and a salt of its conjugate base (ie. HC 2 H 3 O 2 and NaC 2 H 3 O 2 ) HC 2 H 3 O 2 (aq) H 2 O (l) H 3 O (aq) C 2 H 3 O - 2 (aq) OR a weak base and a salt of its conjugate acid (ie. NH 3 and NH 4 Cl). NH 3 (aq) H 2 O (l) OH - (aq) NH 4 (aq) Buffer Calculations Buffer calculations are based on calculations for solutions with a common ion. ex. A solution contains 0.50 M HC 2 H 3 O 2 and 0.50 M NaC 2 H 3 O 2. Calculate the ph of the solution. Determine the resulting ph if mol of NaOH are added to 1.0 L of this buffer. Determine the resulting ph if mol of HCl are added to 1.0 L of this buffer. ex. A solution contains 0.25 M NH 3 and 0.25 M NH 4 Cl. Calculate the ph of the solution. Determine the resulting ph if mol of HCl are added to 1.0 L of this buffer. Determine the resulting ph if mol of NaOH are added to 1.0 L of this buffer.

11 The Henderson-Hasselbalch Equation Determining the ph of any solution containing a weak acid and its conjugate base can be simplified using the following equation: ph [ A ] = pka log [ HA where: pk a = -log K a ] Determining the poh of any solution containing a weak base and its conjugate acid can be simplified using the following equation: poh [ BH ] = pkb log [ B] where: pk b = -log K b Preparing Buffers ex. Calculate the ph of a buffer made by adding 0.50 g of sodium fluoride to 200 ml of 0.10 M hydrofluoric acid. Assume the volume of the solution remains constant. ex. Determine the mass of methylammonium chloride (CH 3 NH 3 Cl) that should be added to 300 ml of 0.20 M methylamine (CH 3 NH 2 ) in order to make a buffer with a ph of Assume the volume of the solution remains constant.

12 Titrations titration: an analytical procedure in which the concentration of a known solution is used to determine the concentration of an unknown solution (used for acids and bases, oxidation and reduction, and solubility, etc) Acid Base Titrations Acid base titrations involve neutralization reactions (recall, acids and bases react to form salt and water) For an experiment, determine the volume of a solution with known concentration that is required to titrate (exactly react with) a given volume of solution with an unknown concentration. The solution with known concentration is called the titrant. The solution with the unknown concentration is called the analyte. Usually several trials are performed and the average volume is used for calculations Chemical indicators (ie. phenolphthalein or bromothymol blue) can be used to monitor when the reaction is complete and has reached its equivalence point. Neutralizations Reactions To determine the net ionic equation for a neutralization reaction, the spectator ions must be decided. Only strong acids, strong bases, and soluble salts will dissociate completely to form ions. Weak acids, weak bases, and insoluble salts will not significantly dissociate. ex. Write the net ion equation for the following neutralization reactions. (1) HBr KOH (2) HF NaOH (3) HNO 3 NH 3 (4) H 2 SO 4 LiOH Titration Calculations ex. What is the volume of 0.50 M HCl required to titrate ml of 0.25 M NH 3? ex. What is the concentration if ml of HF requires ml of 0.10 M NaOH to titrate?

13 Titration Curve Calculations Strong Acid/ Strong Base ex. A M solution of NaOH is used to titrate ml of a M HCl solution. Determine the ph at the following intervals in the experiment. (1) No NaOH has been added ph = (5) ml of NaOH has been added ph = (2) ml of NaOH has been added ph = (6) ml of NaOH has been added ph = (3) ml of NaOH has been added ph = 1.48 (4) ml of NaOH has been added ph = 1.85 (7) ml of NaOH has been added ph = (8) ml of NaOH has been added ph = (9) ml of NaOH has been added. ph = Strong Acid Strong Base Titration ph Volume NaOH added (ml)

14 Strong Acid/Weak Base ex. A M solution of HCl is used to titrate ml of a M NH 3 solution. Determine the ph at the following intervals in the experiment. (1) No HCl has been added ph = (5) ml of HCl has been added ph = (2) ml of HCl has been added ph = (6) ml of HCl has been added ph = (3) ml of HCl has been added ph = 9.26 (4) ml of HCl has been added ph = 8.77 (7) ml of HCl has been added ph = 1.70 (8) ml of HCl has been added ph = 1.56 (9) ml of HCl has been added. ph = 1.30 Strong Acid Weak Base Titration ph Volume HCl added (ml)

15 Summary of Titration Curves Titration Initial ph nearing the equivalence point Change in ph around equivalence point At the equivalence point Final ph Titration Curve Strong Acid added to Strong Base very high steady decrease extreme ph change ph = 7 very low Strong Base added to Strong Acid very low steady increase extreme ph change ph = 7 very high Strong Acid added to Weak Base medium high more gradual decrease (buffer region) more moderate ph change ph<7 very low Strong Base added to Weak Acid medium low more gradual increase (buffer region) more moderate ph change ph>7 very high Acid/Base Indicators The equivalence point in an acid base titration is monitored using an acid base indicator. An indicator is a weak acid that changes colour depending on whether it is in its acidic form or its basic form. ex. Hind (aq) H 2 O (l) acidic form H 3 O (aq) Ind - (aq) basic form The point in a titration at which the indicator changes colour is called the endpoint. Every indicator changes colour at a specific ph value called its pk a value. An indicator is selected for a given titration because its pk a value is close to the ph at the equivalence point. ex. Bromothymol blue is yellow in its acidic form and blue in its basic form. Bromthymol blue has a pk a value of 7.0 and would be useful for a titration with an equivalence point around 7 (ie. a strong acid and a strong base) ex. Phenolphthalein is colourless in its acidic form and bright pink in its basic form. Phenolphthalein has a pk a value of 9.3 would be useful for a titration with an equivalence point around 9 (ie. a weak acid and a strong base) ex. Methyl red is yellow in its acidic fomr and red in its basic form. Methyl red has a pk a value of 5.1 and would be useful for a titration with an equivalence point around 5 (ie. a strong acid and a weak base)

16 Diprotic Titration If a strong base is titrated with a weak diprotic acid, there will be two equivalence points as shown in the following titration curve. Lewis Acid Base Model The Lewis acid base model provides the most general description of an acid or base. Lewis Acid: electron acceptor Lewis Base: electron donator Some common Lewis Acid-Base reactions: (1) Boron compounds (such as BH 3 and BF 3 ) with group five element compounds (usually nitrogen or phosphorus, such as NH 3 or PH 3 ) ex. BF 3 NH 3 BF 3 NH 3 The electrons are accepted by the lewis acid, BF 3 and donated by the lewis base, NH 3. (Generally, the boron compound acts like the lewis acid and the group five element compound acts like the lewis base._ (2) metallic oxide water base ex. BaO H 2 O Ba(OH) 2 The metallic oxide is acting like a lewis base and water is acting like a lewis acid. (3) non-metallic oxide water acid ex. CO 2 H 2 O H 2 CO 3 The non-metallic oxide is acting like a lewis acid and water is acting like a lewis base. (4) metallic oxide non-metallic oxide salt ex. BaO CO 2 BaCO 3 The metallic oxide is acting like a lewis base and the non-metallic oxide is acting like a lewis acid (5) formation of a complex ion. ex. Al 3 6Cl - [AlCl 6 ] 3- ex. Cu 2 4NH 3 [Cu(NH 3 ) 4 ] 2 The metals are acting like lewis acids and the ligands are acting like lewis bases

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